Let us assume a flat FRW metric

$$ds^2 = -dt^2 + a(t)^2(dx^2+dy^2+dz^2)$$

Imagine a lightbeam traveling in the x-direction. It travels on a null geodesic with $ds=0$ so that its path obeys the relation

$$dt = a(t) dx$$

Let us assume that $a=1$ at the present cosmological time.

Thus presently in an interval of cosmological time $dt$ light travels a proper distance of $dx$.

Now imagine a future time at which $a=2$.

Thus in the future in the same interval of cosmological time $dt$ a light beam will travel a proper distance of $2dx$.

Is this correct?

It seems that if an observer uses cosmological time as his time measure then the local speed of light will change with the age of the Universe.

However if he uses conformal time $\tau$ such that an interval of conformal time $d\tau$ is given by

$$d\tau = \frac{dt}{a}$$

then while the proper distance the light travels increases in proportion to $a$ the number of time intervals will also be proportional to $a$ (because their size $d\tau$ is inversely proportional to $a$).

Thus the local speed of light will remain constant if one uses conformal time rather than cosmological time (as it should).

  • $\begingroup$ Huh, how is conformal time in this context defined generally? I only know about proper time ... $\endgroup$
    – Dilaton
    Commented Jun 13, 2013 at 9:57
  • $\begingroup$ One can rewrite the FRW metric in conformal co-ordinates ($\eta$,$x$,$y$,$z$) giving $ds^2=a(\eta)^2(-d\eta^2+dx^2+dy^2+dz^2)$ where the conformal time $\eta=\int \frac{dt}{a(t)}$. My hypothesis is that an inertial observer at rest in an expanding universe will measure conformal time. However most people think such an observer will measure his proper time which in this situation they say is equal to cosmological time. $\endgroup$ Commented Jun 13, 2013 at 11:26

1 Answer 1


No, light always travels by the speed $c$ when measured as the proper distance over proper time in any frame that locally resembles an inertial one. So in the $c=1$ units, $dx_{\rm proper}/dt_{\rm proper}$ is always equal to one, never two.

The expression $ds$ is the proper time (when a time-like interval is substituted to the formula for $ds^2$) while $dt,dx,dy,dz$ are changes of the coordinates that can't be automatically interpreted as proper times and distances.

The proper time may be identified with the cosmological time for static observers, $dt=dt_{\rm proper}$, but the proper distance is not $dx$, i.e. $dx\neq d\ell_{\rm proper}$. Instead, the proper distance is, again, $ds$ when a spacelike interval is substituted to the formula for $ds^2$.

So for this Ansatz, the proper length of a small spatial interval $dx$ is $a(x)dx$ i.e. $2\,dx$ for your future choice of $a(t)$. But the speed is calculated as the proper distance over proper time, $$ v = \frac{d\ell_{\rm proper}}{dt_{\rm proper}} = \frac{a(t)dx}{dt} = 1.$$ This ratio is equal to one because $$ \frac{a(t)dx}{dt} = 1 \,\Leftrightarrow \, dt-a(x)dx = 0 \,\Leftrightarrow\, ds^2=0 $$ it follows from the trajectory's being null, light-like, $ds^2=0$.

You apparently calculated the incorrect speed $v=a(t)=2$ as $dx/dt$, the ratio of the changes of the coordinates, but the coordinates can't be interpreted as proper distances and times. In particular, this identification is wrong for the spatial distances. Proper distances are $a(t)dx$, not $dx$ itself, and that's why the ratio of proper distances and proper times – and that's what defines the speed – is equal to $c=1$.

  • 1
    $\begingroup$ Nice clarification of these issues, I like this too :-) $\endgroup$
    – Dilaton
    Commented Jun 13, 2013 at 11:28

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